ps3sol_6243_2003

ps3sol_6243_2003 - Massachusetts Institute of Technology...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science 6.243j (Fall 2003): DYNAMICS OF NONLINEAR SYSTEMS by A. Megretski Problem Set 3 Solutions 1 Problem 3.1 Find out which of the functions V : R 2 R , 2 (a) V ( x 1 , x 2 ) = x 2 + x 2 ; 1 (b) V ( x 1 , x 2 ) = | x 1 | + | x 2 | ; (c) V ( x 1 , x 2 ) = max | x 1 | , | x 2 | ; are valid Lyapunov functions for the systems (1) x ˙ 1 = x 1 + ( x 1 + x 2 ) 3 , x ˙ 2 = x 2 ( x 1 + x 2 ) 3 ; 2 2 2 2 (2) x ˙ 1 = x 2 x 1 ( x 1 + x 2 ), ˙ x 2 = x 1 x 2 ( x 1 + x 2 ); (3) x ˙ 1 = x 2 | x 1 | , x ˙ 2 = x 1 | x 2 | . The answer is: (b) is a Lyapunov function for system (3) - and no other valid pairs System/Lyapunov function in the list. Please note that, when we say that a Lyapunov function V is defined on a set U , then we expect that V ( x ( t )) should non-increase along all system trajectories in U . In the formulation of Problem 3.1, V is said to be defined on the whole phase space R 2 . Therefore, V ( x ( t )) must be non-increasing along all system trajectories, in order for V to be a valid Lyapunov function. To show that (b) is a valid Lyapunov function for (3), note first that system (3) is defined by an ODE with a Lipschitz right side, and hence has the uniqueness of solutions property. Now, every point ( x 1 , x 2 ) R 2 with x 1 = 0 or x 2 = 0 is an equilibrium of (3). Hence V is automatically valid at those points. At every other point in R 2 , V is 1 Version of October 10, 2003
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
2 differentiable, with dV/dx = [sgn( x 1 ); sgn( x 2 )] being the derivative. Hence V ( x ) f ( x ) = x 1 x 2 x 1 x 2 = 0 at every such point, which proves that V ( x ( t )) is non-increasing (and non-decreasing either) along all non-equilibrium trajectories. Below we list the “reasons” why no other pair yields a valid Lyapunov function. Of course, there are many other ways to show that. For system (1) at x = (2 , 0), we have x ˙ 1 > 0, x ˙ 2 < 0, hence both | x 1 | and | x 2 | are increasing along system trajectories in a neigborhood of x = (2 , 0). Since all Lyapunov function candidates (a)-(c) increase when both | x 1 | and | x 2 | increase, (a)-(c) are not valid Lyapunov functions for system (1). For
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 11/07/2011 for the course AERO 16.36 taught by Professor Alexandremegretski during the Spring '09 term at MIT.

Page1 / 6

ps3sol_6243_2003 - Massachusetts Institute of Technology...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online